Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12   1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Ex 6.1

Transcript

Ex 6.1, 15 The total cost C(𝑥) in Rupees associated with the production of 𝑥 units of an item is given by 𝐶(𝑥) = 0.007𝑥^3 – 0.003𝑥2 + 15𝑥 + 4000. Find the marginal cost when 17 units are produced. Since Marginal Cost is Rate of change in Total Cost w.r.t No of units produced Let MC be the marginal cost & C (𝑥) is the total cost & 𝑥 be the number of units produced So, MC = 𝑑(𝐶(𝑥))/𝑑𝑥 …(1) It is Given that total cost C(𝑥)=0.007𝑥^3−0.003𝑥^2+15𝑥+4000 & we need to find Marginal cost when 17 units produced i.e. MC at 𝑥=17 Now MC = 𝑑(𝐶(𝑥))/𝑥 MC = 𝑑(0.007𝑥^3 − 0.003𝑥^2 + 15𝑥 + 4000)/𝑑𝑥 MC = 𝑑(0.007𝑥^3 )/𝑑𝑥 − 𝑑(0.003𝑥^2 )/𝑑𝑥+ 𝑑(15𝑥)/𝑑𝑥+ 𝑑(4000)/𝑑𝑥 MC = 0.007 𝑑(𝑥^3 )/𝑑𝑥 −0.003 𝑑(𝑥^2 )/𝑑𝑥 +15𝑑(𝑥)/𝑑𝑥+0 MC = 0.007 × 3𝑥^2−0.003 ×2𝑥+15 MC = 0.021𝑥^2−0.006𝑥+15 We need to find MC when 𝑥=17 Putting 𝑥=17 MC = 0.021(17)^2−0.006(17)+15 MC = 6.069 – 0.102 + 15 MC = 20.967 Hence the Required Marginal cost is Rs. 20.967 when 𝑥=17

Ex 6.1 